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h1a - MATH 4171 Graph Theory Homework Set I Solutions...

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MATH 4171 Graph Theory SPRING 2006 Homework Set I - Solutions 1. Describe a real world problem (preferable from your own field) that can be modeled by graphs. Specify vertices, edges, and the incidence relation for your graph. ... 2. Is there a simple graph on 2006 vertices for which no two vertices have the same degree? You need to justify your conclusion. Answer: No. Justification. Suppose there was such a graph G = ( V, E ). Then for each v V , since G is simple, v does not join to itself and there is at most one edge between v and any other vertex. It follows that 0 d G ( v ) 2005, for all v V . Since G has 2006 vertices and there are only 2006 choices for the degrees, the choice of G implies that very number in { 0 , 1 , 2 , ..., 2005 } would be the degree of some vertex. In particular, G has a vertex x with d G ( x ) = 0 and a vertex y with d G ( y ) = 2005. Clearly, this is impossible since either xy E , which would imply d G ( x ) 1 and would contradict d G ( x ) = 0, or xy E , which would imply d G ( y ) 2004 and would contradict d G ( y ) = 2005.
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